Given relation is f(xy)=f(x).f(y)∀x,y∈R .....(i)
On putting x=y=0, we get f(0)=f2(0)⇒f(0)=0,1 but
f(0)=0
⇒f(0)=1,
Now if we put y=0 in (i), we get
f(x)=1
Hence dxdy=1⇒y=x+c
⇒y=x+1 (sincey(0)=1)
⇒y(41)+y(43)
=(41+1)+(43+1)=3.
JEE Main 2019 — Mathematics Algebra
Let f:[0,1]→R be such that f(xy)=f(x).f(y), for all x,y∈[0,1], and f(0)=0. If y=y(x) satisfies the differential equation, dxdy=f(x) with y(0)=1 then y(41)+y(43) is equal to:
Held on 9 Jan 2019 · Verified 6 Jul 2026.
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